Question

If shortest wavelength of hydrogen atom in Lyman series is $x$, then longest wavelength in Balmer series of $He^{+}$ is:

• A. $\frac{9x}{5}$
• B. $\frac{36x}{5}$
• C. $\frac{x}{4}$
• D. $\frac{5x}{9}$

Hint:

Use the Rydberg formula $\frac{1}{\lambda} = RZ^2(\frac{1}{n_1^2} - \frac{1}{n_2^2})$. Shortest wavelength corresponds to $n_2 = \infty$, and longest wavelength corresponds to the first line of the series.

Answer 1

The Correct Option is A

Let's use the energy relation approach where Energy $\Delta E \propto Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$ and $\lambda = \frac{hc}{\Delta E}$, meaning $\lambda \propto \frac{1}{Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})}$.

Transition 1: Shortest wavelength in Lyman series for Hydrogen ($H$, $Z=1$).
Shortest wavelength $\implies$ Maximum energy $\implies n_1 = 1, n_2 = \infty$.
$$\Delta E_1 \propto 1^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = 1$$
Given wavelength is $x$, so $x \propto \frac{1}{1}$.

Transition 2: Longest wavelength in Balmer series for Helium ion ($He^+$, $Z=2$).
Longest wavelength $\implies$ Minimum energy $\implies n_1 = 2, n_2 = 3$.
$$\Delta E_2 \propto 2^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4 \left( \frac{1}{4} - \frac{1}{9} \right) = 4 \left( \frac{5}{36} \right) = \frac{5}{9}$$
Let the required wavelength be $\lambda'$. Then $\lambda' \propto \frac{1}{5/9} = \frac{9}{5}$.

Step 3: Comparing the two proportionalities.
$$\frac{\lambda'}{x} = \frac{9/5}{1} = \frac{9}{5}$$
$$\lambda' = \frac{9x}{5}$$
This confirms that the longest wavelength in the Balmer series of $He^{+}$ is $1.8$ times the shortest wavelength of the Lyman series of Hydrogen.